The two lines x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' will be perpendicular, if and on

Q: Why is the direction ratio for $y$ equal to 1? It feels like we are just guessing.

Great observation! When we isolate $y$, we are essentially looking at the expression $\frac{y}{1}$. In the symmetric form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$, the denominator $m$ represents the direction ratio along the $y$-axis. Since $y$ is just $y$, its coefficient is 1, making the direction ratio 1. It’s not a guess; it’s the mathematical identity of the line's orientation!

Q: Do these lines have to intersect to be perpendicular?

That is a brilliant question! In 3D geometry, two lines are considered perpendicular if the angle between their direction vectors is $90^{\circ}$. They do not necessarily have to intersect (they could be skew lines). The condition $\vec{v_1} \cdot \vec{v_2} = 0$ holds true regardless of whether they intersect or are skew. We are checking the orientation of the lines, not their position.

Q: Is there a faster way without converting to symmetric form?

Yes, there is! You could find the direction vector by taking the cross product of the normal vectors of the two planes that define each line. For line $L_1$, the planes are $x - ay - b = 0$ and $z - cy - d = 0$. The direction vector is $\vec{n_1} \times \vec{n_2}$. However, for this specific problem, the algebraic conversion is much faster and less prone to calculation errors. Always choose the path that minimizes complexity!

Question

The two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ will be perpendicular, if and only if

Options:

(A) $aa' + cc' + 1 = 0$

(B) $aa' + bb' + cc' + 1 = 0$

(C) $aa' + bb' + cc' = 0$

(D) $(a + a')(b + b') + (c + c') = 0$

Sigma Percentile · Accepted Answer

### The Beauty of the Third Dimension

Welcome, future engineers. Today, we are stepping into the realm of 3D geometry.

When you first see equations like $x = ay + b$ and $z = cy + d$, your instinct might be to panic. They do not look like the standard line equations you memorized in school.

However, mathematics is not about memorizing forms; it is about understanding the soul of the geometry. These equations are not obstacles; they are simply a different language describing the same beautiful, straight path in space.

### The Asymmetric Trap

Let’s pause and visualize. In 3D space, a single equation like $x = ay + b$ does not describe a line; it describes a plane because it lacks a constraint on the $z$-variable.

When we provide two such equations, $x = ay + b$ and $z = cy + d$, we are defining the intersection of two planes. When two planes intersect, they create a line.

This is the 'asymmetric form' of a line. To find the angle between two lines, we need their direction ratios. We must translate this asymmetric language into the universal symmetric form:

$$
\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}
$$

### The Transformation

Imagine you are holding a tangled knot of string. To understand its orientation, you must straighten it out. That is exactly what we are doing algebraically.

Given $x = ay + b$, we isolate $y$ to get $\frac{x - b}{a} = y$. Similarly, for the second equation $z = cy + d$, we obtain $\frac{z - d}{c} = y$.

Since both expressions are equal to $y$, they must be equal to each other. We can write this as a unified chain:

$$
\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}
$$

The denominators $a, 1,$ and $c$ are the direction ratios of our line $L_1$. We have successfully extracted the DNA of the line from its asymmetric shell. We repeat this process for line $L_2$, yielding direction ratios of $a', 1,$ and $c'$.

### The Perpendicularity Condition

We now have two lines with direction vectors $\vec{v_1} = (a, 1, c)$ and $\vec{v_2} = (a', 1, c')$. We want them to be perpendicular.

In the language of vectors, perpendicularity is synonymous with the dot product being zero. Because the dot product is defined as $\vec{v_1} \cdot \vec{v_2} = |\vec{v_1}| |\vec{v_2}| \cos(	heta)$, setting $	heta = 90^{\circ}$ forces the dot product to vanish.

We calculate the dot product:

$$
(a)(a') + (1)(1) + (c)(c') = 0
$$

This simplifies elegantly to the final condition:

$$
aa' + cc' + 1 = 0
$$

### Conclusion

Look at what we have achieved. We started with two intimidating, asymmetric equations, and through the power of algebraic manipulation and geometric insight, we derived a simple, elegant condition for perpendicularity.

This is the essence of JEE Advanced mathematics. It is not about brute force; it is about seeing the structure beneath the surface. You have the tools and the logic to conquer these 3D problems with confidence.

Animated Solution for Mathematics - Three Dimensional Geometry: The two lines $x = a y + b, z = cy + d$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ will be perpendicular, if and only if

Select Answer:

Visualized Solution

Visualizing the Lines in 3D

The Strategy: Symmetric Form

Converting $L_{1}$ to Symmetric Form (Part 1)

Converting $L_{1}$ to Symmetric Form (Part 2)

Direction Ratios of $L_{1}$

Converting $L_{2}$ to Symmetric Form

Direction Ratios of $L_{2}$

Condition for Perpendicularity

Applying the Condition

Final Result

The Sigma Insight: Angle Between Two Lines

The Beauty of the Third Dimension

The Asymmetric Trap

The Transformation

The Perpendicularity Condition

Conclusion

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d and x=a′y+b′,z=c′y+d′ will be perpendicular, if and only if